Question

Show that the radius of a tetrahedral void is 0.225 times the radius of the sphere forming a close-packed structure.

Calculate the radii of tetrahedral voids in terms of the radius of the spheres forming close-packed structures.

Answer 1

Consider a tetrahedral void formed by four spheres in a close-packed arrangement. The four spheres of radius R occupy the four corners of a regular tetrahedron ABCD.

Let R be the radius of each sphere and r be the radius of the tetrahedral void. A tetrahedral void is just the right size for the sphere G, which has a radius of r.

Assume that the edge of the cube has a length of a. The face diagonal AC of the right-angled triangle ACE is determined by the following:

AC = `sqrt(AE^2 + EC^2)`

= `sqrt(a^2 + a^2)`

= `sqrt(2)a`

Face diagonal AC = R + R = 2R because the spheres A and C do, in fact, touch one another. Therefore,

2R = `sqrt(2)a`

R = `1/sqrt(2a)`   ...(i)

The body diagonal AF is obtained from the right angled triangle ACF by

AF = `sqrt(AC^2 + CF^2)`

= `sqrt((sqrt2a)^2 + a^2)`

= `sqrt(3)a`   ...(ii)

Since the sphere G sitting in the void touches spheres A and F,

AF = R + 2r + R

= 2R + 2r   ...(iii)

From eqs. (ii) and (iii), we have

2R + 2r = `sqrt(3)a`

R + r = `(sqrt3a)/2`   ...(iv)

Dividing eq. (iv) by eq. (i), we have

∴ `(R + r)/R = (sqrt(3a)/2)/(1/(sqrt2a)) = sqrt(3)/sqrt(2)`

`1 + (r)/R = sqrt(3)/sqrt(2)`

∴ `r/R = sqrt(3)/sqrt(2) - 1`

= `(sqrt(3) - sqrt(2))/sqrt(2)`

= `(1.732 - 1.414)/1.414`

= 0.225

∴ r = 0.225R

The radius of a tetrahedral void is approximately 0.225 times the radius of the sphere forming a close-packed structure.